MHB Why Are Inverses Unique in a Category?

Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading Steve Awodey's book: Category Theory (Second Edition) and am focused on Section 1.5 Isomorphisms ...

I need some further help in order to fully understand some further aspects of Definition 1.3, Page 12, including some remarks Awodey makes after the text of the definition ... ...

The start of Section 1.5, including Definition 1.3 ... reads as follows:View attachment 8354In the text of Definition 1.3 we read the following:

" ... ... Since inverses are unique (proof!), we write $$g = f^{-1}$$. ... ... "Can someone please demonstrate a rigorous proof that in a category, inverses are unique ... ?Help will be appreciated ...

Peter
 
Physics news on Phys.org
You have a isomorphism $f:A \rightarrow B$ in a category $C$

Suppose, there are two arrows $g,h:B \rightarrow A$ such that

$fg=1_B$, $gf=1_A$, $fh=1_B$, and $hf=1_A$

in other word, each is an inverse of $f$, we have to prove that $g=h$

Consider $hfg:B \rightarrow A \rightarrow B \rightarrow A$

On one hand, we have $hfg=h(fg)=h1_B=h$

On the other hand, we have $hfg= \cdots$, can you finish this ?
 
Last edited:
steenis said:
You have a isomorphism $f:A \rightarrow B$ in a category $C$

Suppose, there are two arrows $g,h:B \rightarrow A$ such that

$fg=1_B$, $gf=1_A$, $fh=1_B$, and $hf=1_A$

in other word, each is an inverse of $f$, we have to prove that $g=h$

Consider $hfg:B \rightarrow A\rightarrow B$

On one hand, we have $hfg=h(fg)=h1_B=h$

On the other hand, we have $hfg= \cdots$, can you finish this ?
Thanks Steenis ...

Hmm ... easy when you see how ... :) ...

We have ...

$$hfg = h(fg) = h 1_B = h$$ ... ... ... ... ... (1)

and

$$hfg = (hf)g = 1_A g = g$$ ... ... ... ... ... (2) ... so it follows that ...

(1) (2) $$\Longrightarrow g = h$$Hope that is correct ...Thanks again Steenis ...

Peter
 
Yes that is correct

You did these things before, for instance, in Group Theory

In post #2, I meant $hfg:B \rightarrow A \rightarrow B \rightarrow A$
Already corrected
 
Last edited:
Thread 'Derivation of equations of stress tensor transformation'
Hello ! I derived equations of stress tensor 2D transformation. Some details: I have plane ABCD in two cases (see top on the pic) and I know tensor components for case 1 only. Only plane ABCD rotate in two cases (top of the picture) but not coordinate system. Coordinate system rotates only on the bottom of picture. I want to obtain expression that connects tensor for case 1 and tensor for case 2. My attempt: Are these equations correct? Is there more easier expression for stress tensor...
Back
Top